How to calculate values for -- / ++

Question

I am trying to get the value for the last equation

int a = 0, b = 0, c = 0, x = 0, y = 0, z = 0;

a = b++ + ++c;
printf("a=%d\n", a);

x = y + 1 + ++z;
printf("x=%d\t", x);

printf("b=%d\t", --b);
printf("b=%d\t", b++);
printf("c=%d\t", c+1);
printf("c=%d\t", 2-c);

whats the value of the last equation and why? how do I calculate it

As ever, @Vlad, you're a few seconds ahead of me! Must be a time-zone issue? — Adrian Mole, Oct 02 '19 at 10:58

Vlad from Moscow · Answer 1 · 2019-10-02T11:05:42.483

These statements

printf("c=%d\t", c+1);
printf("c=%d\t", 2-c);

do not change the value of the variable c.

The variable was changed only in this statement

a = b++ + ++c;
          ^^^

If you want to change the variable c in the calls of printf then you should write at least

printf("c=%d\t", c = c+1);
printf("c=%d\t", c = 2-c);

Then the output will look like

a=1
x=2 b=0 b=0 c=2 c=0
                ^^^

Without these changes the output is

a=1
x=2 b=0 b=0 c=2 c=1 
                ^^^

How to calculate values for -- / ++

1 Answers1